Left and Right Vector bundles I am reading a paper that starts talking about 'left vector bundles' and I'm having trouble figuring out what they mean. The specific setup is as follows:
A quarternionic line bundle $L$ over manifold $M$ is a real smooth rank 4 vector bundle with fibers 1-dimensional quaternionic right vector spaces (*). 
A complex quarternionic vector bundle is a pair $(L,J)$ with a quaternionic linear endomorphism $J$ such that $J^2=-1$.
A complex quarternionic vector bundle is thus a rank 2 left complex vector bundle whose complex structure is compatible with the right quaternionic structure (**).
(*) Is a right vector space a vector space $V$ with scalar multiplication only defined on the right? So $V \times F \to V$ for field $F$  but $F \times V \to V$ is not defined? 
(**) If what I say above is correct, what does the compatibility mean? If this complex quaternionic vector bundle is left, then the compatability means
$a(QJ)=(aQ)J,~a\in F, Q\in \mathbb{H}$
or something? Or does the "rank 2" say the compatability is something like
$\mathbb{H}\oplus J\mathbb{H}=\mathbb{H}J\oplus \mathbb{H}$
The paper I am reading is "Quaternionic Analysis on Riemann Surfaces and Differential Geometry" by Pedit and Pinkall (1998). It seems like my confusion is not related at all to the quaternion structure, would apply to any vector space. Also, I have found some other references to the left- and right- vector bundles in double vector bundles, but that also seems not related to this. Does anyone have any clarity?
 A: 
(*) Is a right vector space a vector space V with scalar multiplication only defined on the right? So V×F→V for field F but F×V→V is not defined?

This is correct. The main point is the associativity rule: if $v$ is a vector and $a,b$ scalars, then in a left-vector space we have
$$ a \cdot (b \cdot v) = (a \cdot b) \cdot v$$
and in a right-vector space we have
$$ v \cdot (a \cdot b) = (v \cdot a) \cdot b.$$
One could always just swap the order of the factors to write scalar multiplication on the left in a right-vector space, but that is potentially very confusing, because the associativity rule would be
$$\color{red}{ b \cdot (a \cdot v) = (a \cdot b) \cdot v.}$$
I've colored this equation red because it's a bad idea! Among other things, it would mean that we are not allowed to write $abv$, because the two different interpretations $(ab)v$ and $a(bv)$ can be different.
However, for a commutative field (like the complex numbers), $ab = ba$ so we don't have to develop separate notions of left and right vector spaces in that context.
It is possible to talk about a left-$F$ right-$E$ vector space, where $F$ and $E$ are skew fields: this is a vector space that is a left $F$-vector space and a right $E$-vector space that are "compatible": they satisfy an additional associativity constraint:
$$ f \cdot (v \cdot e) = (f \cdot v) \cdot e. $$
Unfortunately, I'm not familiar with your context, so I can't answer your questions directly. In fact, I don't think I've ever done linear algebra over skew fields before -- all of these ideas I'm familiar with from module theory. But, at least, module theory is a generalization: a left vector space over a (skew) field $F$ is the same thing as a left module over $F$ (and the same on the right), so I'm assuming all of the notions you're talking about have the same meaning as the module-theoretic version.
